Find $y^\prime$ if $y=\tanh ^{-1}\ \sqrt{x}$.

Solution We use the Chain Rule with $y=\tanh ^{-1}u$ and $u=\ \sqrt{x}$. Then $$\begin{eqnarray*} y^\prime &=&\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\dfrac{d}{du} \tanh ^{-1}u\cdot \dfrac{d}{dx}\ \sqrt{x}\\ &=&\dfrac{1}{{1-u^{2}}}\cdot \dfrac{1}{2\ \sqrt{x}}=\dfrac{1}{ {1-x}}\cdot \dfrac{1}{2\ \sqrt{x}}=\dfrac{\ \sqrt{x}}{2x(1-x)}\qquad \ {\color{#0066A7}{\hbox{\(u=\ \sqrt{x}\)}}} \nonumber \\ \end{eqnarray*}$$